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b^2-16b+39=0
a = 1; b = -16; c = +39;
Δ = b2-4ac
Δ = -162-4·1·39
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{100}=10$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-10}{2*1}=\frac{6}{2} =3 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+10}{2*1}=\frac{26}{2} =13 $
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